Check the solution to "Solve the initial value problem: (tan⁡(y)−2)dx+(xsec2(y)+1y)dy=0, y(0)=1"

High School
Calculus and Analysis
Calculus 2
Differential equations

tabita57i

Answered question

2021-01-08

Solve the initial value problem: $(\tan (y) - 2) d x + (x \sec^{2} (y) + \frac{1}{y}) d y = 0$ , $y (0) = 1$

Answer & Explanation

Nichole Watt

Skilled2021-01-09Added 100 answers

$(\tan (y) - 2) d x + (x \sec^{2} (y) + \frac{1}{y}) d y = 0$
$\to (\tan (y) d x + x \sec^{2} (y) d y) - 2 d x + (\frac{1}{y}) d y = 0$
$\to d (x \tan (y)) - d (2 x) + d (\ln y) = 0$
$\to d (x \tan (y) - 2 x + \ln y) = 0$
$\to x t a n (y) - 2 x + l n y = c$ ,
x=0, y=1. So, $x \tan (y) - 2 x + \ln y = c$
$\to 0 \cdot \tan (1) - 2 \cdot 0 + \ln (1) = c$
$\to c = \ln (1) = 0$
The general solution of the given intial value problem is $x \tan (y) - 2 x + \ln y = 0$

Jeffrey Jordon

Expert2021-08-11Added 2605 answers

Answer is given below (on video)

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